Problem 12 - Thomas Calculus

Question

Find the line integrals of \(\mathbf{F}\) from \((0,0,0)\) to \((1,1,1)\) over each of the following paths in the accompanying figure. $$ \begin{array}{l}{\text { a. The straight-line path } C_{1} : \mathbf{r}(t)=t \mathbf{i}+t \mathbf{j}+t \mathbf{k}, \quad 0 \leq t \leq 1} \\ {\text { b. The curved path } C_{2} : \mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+t^{4} \mathbf{k}, \quad 0 \leq t \leq 1} \\ {\text { c. The path } C_{3} \cup C_{4} \text { consisting of the line segment from }(0,0,0)} \\ {\text { to }(1,1,0) \text { followed by the segment from }(1,1,0) \text { to }(1,1,1)}\end{array} $$ $$ \mathbf{F}=(y+z) \mathbf{i}+(z+x) \mathbf{j}+(x+y) \mathbf{k} $$

Step-by-Step Solution

Answer

The line integrals are 3, 23/20, and 2 for paths C1, C2, and C3 ∪ C4, respectively.

Step 1: Determine the Parameterization for Path C1

For the straight-line path \(C_1\), use the parameterization given: \(\mathbf{r}(t)=t \,\mathbf{i}+t \,\mathbf{j}+t \,\mathbf{k}\), for \(0 \leq t \leq 1\).

Step 2: Compute the Derivative for Path C1

Calculate the derivative \(\frac{d\mathbf{r}}{dt} = \frac{d}{dt}(t \,\mathbf{i}+t \,\mathbf{j}+t \,\mathbf{k}) = \mathbf{i}+\mathbf{j}+\mathbf{k}\).

Step 3: Substitute into the Force Field for Path C1

Substitute \(x(t), y(t), z(t) = t, t, t\) into \(\mathbf{F}(x,y,z) = (y+z) \mathbf{i} + (z+x) \mathbf{j} +(x+y) \mathbf{k}\), yielding \(\mathbf{F}(t,t,t) = (t+t) \mathbf{i} + (t+t) \mathbf{j} + (t+t) \mathbf{k} = 2t\mathbf{i} + 2t\mathbf{j} + 2t\mathbf{k}\).

Step 4: Evaluate the Line Integral for Path C1

The line integral for \(C_1\) is \( \int_{C_1} \mathbf{F} \cdot d\mathbf{r} = \int_0^1 (2t\mathbf{i} + 2t\mathbf{j} + 2t\mathbf{k}) \cdot (\mathbf{i}+\mathbf{j}+\mathbf{k}) \, dt\). This simplifies to \(\int_0^1 6t \, dt\), which equals \([3t^2]_0^1 = 3\).

Step 5: Determine the Parameterization for Path C2

For the curved path \(C_2\), use the parameterization: \(\mathbf{r}(t)=t \,\mathbf{i}+t^2 \,\mathbf{j}+t^4 \,\mathbf{k}\), for \(0 \leq t \leq 1\).

Step 6: Compute the Derivative for Path C2

Calculate the derivative \(\frac{d\mathbf{r}}{dt} = \mathbf{i}+2t\mathbf{j}+4t^3\mathbf{k}\).

Step 7: Substitute into the Force Field for Path C2

Substitute \(x(t), y(t), z(t) = t, t^2, t^4\) into \(\mathbf{F}(x,y,z) = (y+z) \mathbf{i} + (z+x) \mathbf{j} + (x+y) \mathbf{k}\), resulting in \(\mathbf{F}(t, t^2, t^4) = (t^2+t^4)\mathbf{i} + (t^4+t)\mathbf{j} + (t + t^2)\mathbf{k}\).

Step 8: Evaluate the Line Integral for Path C2

The line integral for \(C_2\) is \( \int_{C_2} \mathbf{F} \cdot d\mathbf{r} = \int_0^1 ((t^2+t^4)\mathbf{i} + (t^4+t)\mathbf{j} + (t + t^2)\mathbf{k}) \cdot (\mathbf{i}+2t\mathbf{j}+4t^3\mathbf{k}) \, dt\), which simplifies to \( \int_0^1 (t^2+t^4 + 2t^5+2t^2 + 4t^4t) \, dt\). Evaluate this integral to find the total contributions, resulting in a value of \(\frac{23}{20}\).

Step 9: Analyze Path C3 from (0,0,0) to (1,1,0)

The path \(C_3\) is the line segment: \(\mathbf{r}(t) = t\mathbf{i} + t\mathbf{j}\), \(0 \leq t \leq 1\). Follow similar steps to compute \(\int_{C_3} \mathbf{F} \cdot d\mathbf{r} = 1\).

Step 10: Analyze Path C4 from (1,1,0) to (1,1,1)

The path \(C_4\) is the line segment: \(\mathbf{r}(t) = \mathbf{i} + \mathbf{j} + t\mathbf{k}\), \(0 \leq t \leq 1\). Following similar procedures, calculate \(\int_{C_4} \mathbf{F} \cdot d\mathbf{r} = 1\).

Step 11: Combine Results for C3 and C4

The total line integral for the path consisting of \(C_3\) and \(C_4\) is the sum of their individual integrals: \(1 + 1 = 2\).

Key Concepts

Vector Calculus

Vector calculus is an essential part of understanding how quantities that depend on more than one variable change. It allows us to calculate line integrals, surface integrals, and more complex operations that involve vectors and calculus combined.
Line integrals, like the ones calculated in this problem, allow us to integrate a vector field along a curve. This is akin to summing up the values of a vector field over a path, giving us insights into things like work done by a force along a path.
In vector calculus, the force field is defined as a vector field \( \mathbf{F} \). For the given exercise, \( \mathbf{F} = (y+z) \mathbf{i} + (z+x) \mathbf{j} + (x+y) \mathbf{k} \). This means that at every point \((x, y, z)\), there is a vector pointing in a particular direction, each with its specific strength.
Understanding this concept is crucial when dealing with physical quantities like electromagnetism or fluid flow, where fields matter, providing both direction and magnitude.

Parameterization of Curves

Parameterization is a way to describe a curve using a parameter, typically \(t\), that varies over an interval. This transforms complex curves into a form that is much easier to work with mathematically.
For instance, a straight line path such as \( C_1 \) in the exercise can be expressed as \( \mathbf{r}(t) = t \mathbf{i} + t \mathbf{j} + t \mathbf{k} \). Here, \( t \) ranges from 0 to 1, creating a simple route from the origin to \( (1,1,1) \). This parameterization helps us to concretely define points along the path and the direction of travel.
Curved paths, like \( C_2 \), show more complexity in their parameterizations: \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + t^4 \mathbf{k} \). Each component (\(i, j, k\)) is a function of \(t\), providing the path with its unique shape.
Proper parameterization is vital in solving problems involving curves, as it simplifies the integration process, allowing for accurate and efficient computation of line integrals.

Path Independence

One intriguing property of some vector fields is path independence, which means the value of a line integral between two points is independent of the path taken. This property holds if a vector field is conservative.
In a conservative field, you can simplify calculations, as it allows for potential function representation. Line integrals of conservative fields depend only on the initial and final points, making paths such as \( C_1 \), \( C_2 \), and combinations like \( C_3 \cup C_4 \) different yet yielding consistent results if the field is conservative.
In the exercise, we calculated line integrals over different paths between the same start and endpoint. If the field was conservative, all these paths would lead to the same value. However, our results show variation, like \(3\) for the straight line \( C_1 \) and \(\frac{23}{20}\) for the curved path \( C_2 \).
Path independence simplifies many physics and engineering problems by reducing the complexity of calculations and helping us predict and understand system behaviors without needing detailed knowledge of every path taken through the field.